Question

How do you find the area of an equilateral triangle with a 6-inch radius?

Answer 1

Based on this image...

Explanation:

...we can call segment "a" the radius, and if we do that, we calculate the area of the triangle as follows:

Look at the vertical line bisecting the triangle, made up of segments of length "a" and "h"

Now look at the smaller, right triangle towards the bottom left of the equilateral triangle, with sides "a", "h", and "S/2".

If we're taking "a" to be the radius (in our case 6 inches), then

`h/a = sin 30` degrees, so `h/6 = 0.5`, so h = 3 inches. Therefore the height of the equilateral triangle is `6+3 = 9` inches.

The original equilateral triangle in the figure is bisected by the vertical line, making 2 right triangles of height 9 inches and a base of length S/2.

`(S/2)/a = cos 30`, so `S/2 = 0.866 * 6 = 5.196` inches (rounding).

So, the original equilateral triangle in the figure is twice the area of this larger right triangle. A right triangle's area is `1/2b*h`. We have TWO of them, so the total area is `2 * 1/2b * h = 5.196 * 9`

...which works out to:

46.77 square inches (rounding)

GOOD LUCK